A runner runs around a circular track. He completes one lap at a time of t = 314 s at a constant speed of v = 3.1 m/s. t = 314 sv = 3.1 m/sWhat is the radius, r in meters, of the track? What was the runners centripetal acceleration, ac in m/s2, during the run?

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CorrieR176843 CorrieR176843 · Answer 1 · 2022-10-25T16:54:24+02:00

Since the runner completes 1 lap in 314 seconds, and its velocity is 3.1m/s, then the total distance covered in 1 lap is:

[tex]\begin{gathered} d=vt \\ =(3.1\frac{m}{s})(314s) \\ =973.4m \end{gathered}[/tex]

That distance corresponds to the perimeter of the circumference. The perimeter of a circumference with radius r is 2πr. Then:

[tex]\begin{gathered} 2\pi r=d \\ \\ \Rightarrow r=\frac{d}{2\pi} \\ =\frac{973.4m}{2(3.14...)} \\ =154.9...m \end{gathered}[/tex]

The centripetal acceleration of an object in a circular trajectory with radius r and speed v is:

[tex]a_c=\frac{v^2}{r}[/tex]

Replace v=3.1m/s and r=154.9m to find the centripetal acceleration:

[tex]a_c=\frac{(3.1\frac{m}{s})^2}{(154.9m)}=0.062\frac{m}{s^2}[/tex]

Therefore, the radius of the track is approximately 155m and the centripetal acceleration of the runner is approximately 0.062 m/s^2.